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Reduction of a simple distributed loading (correct1)

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  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    View_image_1456412389869.jpg
    I am not yet at the chapter of equations of equilibrium, plus it says the couple moment is not 0, so I assume it just about the loadings and not including the reactive forces and reactive moment.

    2. Relevant equations
    F= 1/2*b*h
    M= 1/2*b*h*d

    3. The attempt at a solution
    So if I call the top one F1 and the bottom loading F2 I got:

    F1+F2=0= -4*b*1/2+2.5*(b+a)*1/2
    solving this i got a=0.6b

    Moment around the of the bar (not A, but opposite side) due to the loadings is (counter clockwise positive):

    M= -F1*center of triangle + F2*center of triangle
    Because you can replace the loading with force F1 and force F2 and its line of action is through the center of the triangle area (1/3* base)
    M= -8= -4*b*1/2*1/3b+2.5*(b+a)*1/2*1/3*(b+a)

    So what am I doing wrong?
    Because M can never be negative with me, plus the answer should be b= 5.625 and a= 1.539.
    But this to me makes no sense, because then F1+F2 is not 0.
    And if I should take the reactive forces into account at A then you can never have still a moment, because then it is not static anymore.
     
  2. jcsd
  3. Feb 25, 2016 #2

    SteamKing

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    I agree with this.
    If your moment convention is negative for CW moments, then you should carry this convention thru your moment calculations.
    The values a = 1.539 and b = 5.625 do not produce a zero net force on the beam with the indicated loadings, so they cannot be correct.

    You never said what you calculated for a and b.
     
  4. Feb 25, 2016 #3

    But I did carry it though my moment equations?
    M=-8 = - F1*d1 + F2* d2
    This is what I did, F1 produces a clockwise moment (thus negative sign) and F2 produces a counterclockwise moment (thus positive).

    I did not get a value for a and b, since - F1*d1 + F2* d2 never get below 0 if I use M= -8= -4*b*1/2*1/3b+2.5*(b+a)*1/2*1/3*(b+a)
     
  5. Feb 25, 2016 #4

    SteamKing

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    I don't know how you can say - F1*d1 + F2* d2 never got below zero because you never showed this to be true and you never worked out a value for a or b.
    It really doesn't matter, since the values of a and b which are the supposed "solutions" are incorrect.
     
  6. Feb 25, 2016 #5
    well i said a=0.6b, you also found that correct now
    M= 4*b*1/2*1/3b+2.5*(b+a)*1/2*1/3*(b+a)
    filling in 0.6b=a gives M= -2/3b^2+ 5/12(b+0.6b)^2
    which gives M= 0.4b^2
    Which can never get below zero. You can graph it or use the D<0 method
     
  7. Feb 25, 2016 #6

    SteamKing

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    I took moments about the fixed end of the beam. In that case, the top moment was CCW and the bottom moment was CW. I got a net CW moment, which allowed me to solve for a and b, but these were different values than the ones which were supposed to be the solutions.
     
  8. Feb 25, 2016 #7

    haruspex

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    Since the downwards forces have an effective line of action to the left of that of the upward forces, the net moment is clearly anticlockwise. You are right, the book is wrong.
     
  9. Feb 25, 2016 #8

    haruspex

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    the OP method is simpler.
     
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